An arrow shot vertically upwards loses its initial speed by 60% in 3 seconds. The maximum height reached by the arrow is ($g=9.8m{s}^{-2}$) ?

The correct option is C $122.5m$

Let initial speed be u, so after $3$ sec it will be,

$u-60\%u=u-\frac{60}{100}u=\frac{2}{5}u=v$

$\therefore v=u+at$

$\therefore \frac{2}{5}u=u-9.8\times 3$ .....($a=g=9.8,t=3$sec)

$\therefore u=4gm/s$

$\therefore v^2=u^2+2at$

$\therefore 0=u^2-2\times 9.8\times H$ .......(at max height $v=0$)

$\therefore H=\frac{u^2}{2\times 9.8}=\frac{4g\times 4g}{2\times 9.8}$

$\therefore$ Maximum height$=H=122.5$m.